EE 369

POWER SYSTEM ANALYSIS

Lecture 8 Transformers, Per Unit Tom Overbye and Ross Baldick

1

Announcements

• For lectures 8 to 10 read Chapter 3  HW 6 is problems 5.2, 5.4, 5.7, 5.9, 5.14, 5.16, 5.19, 5.26, 5.31, 5.32, 5.33, 5.36; case study questions chapter 5 a, b, c, d, is due Thursday, 10/15.

 Homework 7 is 5.8, 5.15, 5.17, 5.24, 5.27, 5.28, 5.29, 5.34, 5.37, 5.38, 5.43, 5.45; due 10/22. 2

• • • •

Transformers Overview

Power systems are characterized by many different voltage levels, ranging from 765 kV down to 240/120 volts. Transformers are used to transfer power between different voltage levels.

The ability to inexpensively change voltage levels is a key advantage of ac systems over dc systems.

In this section we’ll development models for the transformer and discuss various ways of connecting three phase transformers. 3

Ideal Transformer

 First we review the voltage/current relationships for an ideal transformer – no real power losses – magnetic core has infinite permeability – no leakage flux  We’ll define the “primary” side of the transformer as the side that usually receives power from a line etc, and the secondary as the side that usually delivers power to a load etc: – primary is usually the side with the higher voltage, but may be the low voltage side on a generator step up transformer. 4

Ideal Transformer Relationships

Note that

I

2 and

I

2 ’ are in opposite directions Assume we have flux 

m

in magnetic material. Then flux linking coil 1 having

N

1  1 

N

1 

m

, and similarly  2 

v

1 

d

dt

1 

N

1

d

m dt

,

v

2 

d

m

dt v

1

N

1 

v

2

N

2 

V

1

V

2 turns is:

N d

 2 

m

dt

2

N

1

N

2  

N

2

d

m dt a

= turns ratio 5

Current Relationships

To get the current relationships use ampere's law mmf   

H

 '   '    ' Assuming uniform flux density in the core having area , then    

L

A

  '

A

6

  ' . Hence

i

1

i

2 '  

N N

2 1 or

I

Then:

I

2 1  1 and:

a V

1     

a

0 0 1  

a i

1

i

2 

N

2

N

1  1

a

, where

i

2  

i

2 ' 7

Impedance Transformation Example

Example: Calculate the primary voltage and current for an impedance load Z on the secondary

I

2 

V

2

V

1     

a

0 0 1

a

   

V V

2 2

Z

 

V

1 

aV

2

I

1  1

V

2

a Z V

1

I

1   primary referred value of secondary load impedance 8

Real Transformers

• Real transformers – have losses – have leakage flux – have finite permeability of magnetic core • Real power losses – resistance in windings (I 2 R) – core losses due to eddy currents and hysteresis 9

Transformer Core losses

Eddy currents arise because of changing flux in core.

Eddy currents are reduced by laminating the core Hysteresis losses are proportional to area of BH curve and the frequency These losses are reduced by using material with a “thin” BH curve 10

Effect of Leakage Flux

Not all flux is within the transformer core  1  2   

l

1 

l

2  

N

1 

m N

2 

m

, where 

l

1 , where 

l

2 is the coil 1 leakage flux, is the coil 2 leakage flux, Assuming a linear magnetic medium we get 

l

1

v

1 

l

l

2 

L l

1

di

1

dt

N

1

l d

m

2 ' , including winding

dt r v

2  ' 2 

L l

2

di

2 '

dt

N

2

d

m dt

, including resistance .

11

Effect of Finite Core Permeability

Finite core permeability means a non-zero mmf is required to maintain  m  

R

 m , in the core This effect is usually modeled as a magnetizing current

i

1 

R

 m

N

1 

N N

1 2

i

2

i

1 

i

m 

N

2

i

2

N

1 where

i

m 

R

 m

N

1 , modeled by resistance and inductance.

12

Transformer Equivalent Circuit

Using the previous relationships, we can derive an equivalent circuit model for the real transformer This model is further simplified by referring all impedances to the primary side (and approximating by swapping the referred elements and the shunts):

r

2 ' 

x

2 '  2 2

r e x e

r

2 '

x

1 

x

2 ' 13

14

Calculation of Model Parameters

 The parameters of the model are determined based upon: – nameplate data: gives the rated voltages and power – open circuit test: rated voltage is applied to primary with secondary open; measure the primary current and losses (the test may also be done applying the rated voltage to the secondary, calculating the values, then referring the values back to the primary side).

– short circuit test: with secondary shorted, apply (lower than rated) voltage to primary to get rated primary current to flow; measure voltage and losses.

15

Transformer Example

• Example: A single phase, 100 MVA, 200/80 kV transformer has the following test data: – open circuit: 20 amps, with 10 kW losses – short circuit: 30 kV, with 500 kW losses • Determine the model parameters. 16

Transformer Example, cont’d

From the short circuit test

I sc

 100MVA 200kV  500 A,

R e

jX e

 30 kV 500 A  60 

P sc

 2  500 kW 

R e

P sc

/

I

2

sc

 500,000 /(500) 2 Hence

X e

 60 2  2 2  60

From the open circuit test

R c

 (

V

rated ) 2

P oc

 2 (200) (kV) 2 10 kW  4M 

R e

jX e

jX m

V

rated

I oc

 200 kV 20 A  10,000 

X m

,  10,000  17

Residential Distribution Transformers

Single phase transformers are commonly used in residential distribution systems. Most distribution systems are 4 wire, with a multi-grounded, common neutral. 18

Per Unit Calculations

 A key problem in analyzing power systems is the large number of transformers. – It would be very difficult to continually have to refer impedances to the different sides of the transformers  This problem is avoided by a normalization of all variables.

 This normalization is known as per unit analysis. quantity in per unit  actual quantity base value of quantity 19

Per Unit Conversion Procedure, 1

 1. Pick a 1  VA base for the entire system, S B 2. Pick a voltage base for each different voltage level, V B . Voltage bases are related by transformer turns ratios. Voltages are line to neutral.

3. Calculate the impedance base, Z B = (V B ) 2 /S B 4. Calculate the current base, I B = V B /Z B 5. Convert actual values to per unit Note, per unit conversion affects magnitudes, not the angles. Also, per unit quantities no longer have units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts) 20

Per Unit Solution Procedure

1. Convert to per unit (p.u.) (many problems are already in per unit) 2. Solve 3. Convert back to actual as necessary 21

Per Unit Example

Solve for the current, load voltage and load power in the circuit shown below using per unit analysis with an S B of 100 MVA, and voltage bases of 8 kV, 80 kV and 16 kV, respectively. Original Circuit 22

Per Unit Example, cont’d

Z Left B

2  100MVA  0.64

Z Middle B

2  100MVA

Z Right B

2  100MVA  2.56

 Same circuit, with values expressed in per unit. 23

Per Unit Example, cont’d

I V L

  3.91

j

2.327

0.22

 0.22

    30.8

    p.u.

 

S S L G

  * 

V L

2  0.189 p.u.

Z

0.22 30.8

    24

Per Unit Example, cont’d

To convert back to actual values just multiply the per unit values by their per unit base

V L

Actual 

S L

Actual  0.859

  30.8

 16 kV 100 MVA  

S

Actual

G

 0.22 30.8

 100 MVA  13.7

 

I I

Middle

B

Actual Middle   100 MVA  1250 Amps 80 kV 0.22

  30.8

  275   30.8

 25

Three Phase Per Unit

Procedure is very similar to 1  except we use a 3  VA base, and use line to line voltage bases 1. Pick a 3  VA base for the entire system,

S

3

B

 2. Pick a voltage base for each different voltage level, V

B,LL

. Voltages are line to line . 3. Calculate the impedance base

Z B

 2

V S

3 

B

 ( 3

V

3

S

1 

B

) 2  2

V S

1 

B

Exactly the same impedance bases as with single phase using

Three Phase Per Unit, cont'd

4. Calculate the current base, I B

I

3  B 

S

3 

B

 3

S

1 

B

 3

V

3 3

V V S

1 

B

I

1  B Exactly the same current bases as with single phase!

5. Convert actual values to per unit 27

Three Phase Per Unit Example

• Solve for the current, load voltage and load power in the previous circuit, assuming: – a 3  power base of 300 MVA , – and line to line voltage bases of 13.8 kV, 138 kV and 27.6 kV (square root of 3 larger than the 1  example voltages) – the generator is Y-connected so its line to line voltage is 13.8 kV. Convert to per unit as before.

Note the system is exactly the same!

28

Per Unit Example, cont'd

I V L

  3.91

j

2.327

0.22

 0.22

    30.8

    p.u.

 

S S L G

  * 

V L

2  0.189 p.u.

Z

0.22 30.8

    Again, analysis is exactly the same!

29

Per Unit Example, cont'd

Differences appear when we convert back to actual values

V

L Actual 

S

Actual L 

S

Actual G  0.859

  30.8

 27.6 kV  23.8

  300 MVA   300 MVA 

I I

Middle B Actual Middle   300 MVA 3 138 kV  125 0 Amps (same cu rrent !) 0.22

  30 .

8   Amps  275   30.8

 30